Question: A curve is defined by the parametric equations $x=\sqrt[ 3]{t}-6$ and $y=\sqrt{t}+1$. What is $\dfrac{d^2y}{dx^2}$ in terms of $t$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{3}{4\sqrt[ 6]{t}}$ (Choice B) B $\dfrac{3\sqrt[ 6]{t}}{2}$ (Choice C) C $\dfrac{9\sqrt[ 6]{t^5}}{2}$ (Choice D) D $\dfrac{3\sqrt[ 6]{t^5}}{2}$
Solution: We are asked to find the second derivative of a parametric function. Recall that the first derivative of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ is found with the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ Then, the second derivative is found with this following rule: $\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{\dfrac{d}{dt}\left(\dfrac{v'(t)}{u'(t)}\right)}{u'(t)}$ Let's start by finding $\dfrac{dy}{dx}$. $\dfrac{dy}{dx}=\dfrac{3\sqrt[ 6]t}{2}$ Now we can find $\dfrac{d^2y}{dx^2}$. $\begin{aligned} \dfrac{d^2y}{dx^2}&=\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dx}{dt}\right)} \\\\ &=\dfrac{\dfrac{d}{dt}\left(\dfrac{3\sqrt[ 6]t}{2}\right)}{\dfrac{d}{dt}(\sqrt[ 3]{t}-6)} \\\\ &=\dfrac{\left(\dfrac{t^{ -\frac56}}{4}\right)}{\left(\dfrac{t^{ -\frac23}}{3}\right)} \\\\ &={\dfrac{3t^{\left( \frac23-\frac56\right)}}4} \\\\ &=\dfrac{3t^{ -\frac16}}{4} \\\\ &=\dfrac{3}{4\sqrt[ 6]{t}} \end{aligned}$ In conclusion, $\dfrac{d^2y}{dx^2}=\dfrac{3}{4\sqrt[ 6]{t}}$.